Yes, with F(40) (wow), with such a value you can get further, but a little detail, you need over 152319 digits to get there.! exp(Pi*40)^2791 = 0.946548568816 x 10^152319. I took the simplest value enough to compute the classical example of MacMahon. still, I am surprised by the relative simplicity of that algebraic number, Simon Plouffe Le 2011-03-16 10:55, Bill Gosper a écrit :
(With much help from Julian, who says this is good thru p(2791).) F(40) = sqrt(sqrt(2) + 1) * (5^(1/4) + 1)^(3/2) * sqrt(5^(1/4) + sqrt(2)) * sqrt(5) * (sqrt(5) + 1)^(11/4) * (sqrt(10) + 3)^(1/4) * pi^(3/4) * exp( - 5 * pi/3)/(2^(27/16) * GAMMA(1/4)) --rwg By-product: 3/4 (d83) 6 sqrt(2) 5 + (69 sqrt(2) + 12) sqrt(5) =
1/4 102 sqrt(2) 5 + 21 sqrt(2) + 28
(c84) DFLOAT(%);
(d84) 273.402222551711d0 = 273.40222255171d0
(c85) BFLOAT(D83);
(d85) 2.7340222255171092466b2 = 2.7340222255171007641b2